Problem: Simplify the following expression: $y = \dfrac{4x^2+7x- 2}{4x - 1}$
Solution: First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(4)}{(-2)} &=& -8 \\ {a} + {b} &=& &=& {7} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $-8$ and add them together. Remember, since $-8$ is negative, one of the factors must be negative. The factors that add up to ${7}$ will be your ${a}$ and ${b}$ When ${a}$ is ${-1}$ and ${b}$ is ${8}$ $ \begin{eqnarray} {ab} &=& ({-1})({8}) &=& -8 \\ {a} + {b} &=& {-1} + {8} &=& 7 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({4}x^2 {-1}x) + ({8}x {-2}) $ Factor out the common factors: $ x(4x - 1) + 2(4x - 1)$ Now factor out $(4x - 1)$ $ (4x - 1)(x + 2)$ The original expression can therefore be written: $ \dfrac{(4x - 1)(x + 2)}{4x - 1}$ We are dividing by $4x - 1$ , so $4x - 1 \neq 0$ Therefore, $x \neq \frac{1}{4}$ This leaves us with $x + 2; x \neq \frac{1}{4}$.